$\sum_{i\in I} \kappa_i \leq \prod_{i\in I} \lambda_i $ when $\kappa_i < \lambda_i$ for all $i\in I$ and $\kappa_i,\lambda_i$ are cardinals.

Question

I am struggling with this exercise, as the title says, I have to prove that $\sum_{i\in I} \kappa_i \leq \prod_{i\in I} \lambda_i $ when $\kappa_i < \lambda_i$ for all $i\in I$ and $\kappa_i,\lambda_i$ are cardinals.

Now, I started from noticing that $\sum_{i\in I} \kappa_i \leq \sum_{i\in I} \lambda_i$ and I tried to work with the definitions.

$\sum_{i\in I}\lambda_i = |\bigcup_{i\in I}(\text{{$i$}}\times\lambda_i)|$

$\prod_{i\in I}\lambda_i = |\{f:I\to\bigcup_{i\in I}\lambda_i : \forall i\in I, f(i)\in\lambda_i\}|$

So I tried to find an injective function between the two sets $\bigcup_{i\in I}(\text{{$i$}}\times\lambda_i)$ and $\{f:I\to\bigcup_{i\in I}\lambda_i : \forall i\in I, f(i)\in\lambda_i\}$ but I have no idea on how to do that. Is that the way to go? If so can you help me with finding such an injective function? Thank you in advance guys :)

Answer 1

Using $\kappa_i < \lambda_i$ we have that $\lambda_i \setminus \kappa_i$ is non-empty for all $i \in I$. Let $f \in \prod_{i \in I} \lambda_i$ be such that $f(i) \in \lambda_i \setminus \kappa_i$ for all $i \in I$. Now for any $(i, \alpha) \in \sum_{i \in I} \kappa_i$ we define $$ g_{i, \alpha}(j) = \begin{cases} f(j) & \text{if } j \neq i \\ \alpha & \text{if } j = i \end{cases} $$ The assignment $(i, \alpha) \mapsto g_{i,\alpha}$ then is an injection $\sum_{i \in I} \kappa_i \to \prod_{i \in I} \lambda_i$.